Q. 14.2( 15 Votes )

# How much paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square with diagonal 44 cm.

Answer :

AC = BD = 44cm

AO = 44/2 = 22cm

BO = 44/2 = 22cm

In ΔAOB,

AB^{2} = AO^{2} + BO^{2}

⇒ AB^{2} = 22^{2} + 22^{2}

⇒ AB^{2} = 2 × 22^{2}

⇒ AB = 22√ 2 cm

Area of square = (Side)^{2} = (22√2)^{2} = 968 cm^{2}

Area of each triangle (I, II, III, IV) = Area of square /4 = 968 /4 = 242 cm^{2}

Area of lower triangle,

a = 28, b = 28, c = 14

s = (a + b + c)/2

⇒ s = (28 + 28 + 14)/2 = 70/2 = 35.

Area(Δ) = √s(s-a)(s-b)(s-c)

⇒ Area(Δ) = √35(35-28)(35-28)(35-14)

⇒ Area(Δ) = √35×7×7×21

⇒ Area(Δ) = 49√15 = 189.77cm^{2}

Area of Red = Area of IV = 242 cm^{2}

Area of Yellow = Area of I + Area of II = 242 + 242 = 484 cm^{2}

Area of Green = Area of III + Area of lower triangle = 242 + 189.77 = 431.77 cm^{2}

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